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Mathematical olympiad of primary school students' mathematical reverse motion problem
Think like this:

First, consider that the speed of A is 4km/ h and the speed of B is 6km/ h, so there is 24/(4+6)=2. Four hours. It means that it will take more than 2 hours for the two to meet for the first time.

Both of them walk in two stages (taking 1 hour as one stage): A can walk: 4*2=8KM, and the time is 1*2 hours +5*2 minutes =2 hours 10 minute; B can walk: 6*2= 12KM, time: (50+ 10)*2= 120 minutes =2 hours.

Obviously, it takes 2 hours for A to go to the third paragraph, 10 minutes, while B only needs 2 hours. That is, Party B goes first 10 minute, and Party B can go 10 minute:

6* 10/60= 1KM, and then there is 24-8- 12- 1=3KM left. 3/(4+6)=3/ 10 hour = 18 minutes.

So the time when they met for the first time was: 2 hours 10 minutes+18 minutes =2 hours and 28 minutes.